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steposvetlana [31]
3 years ago
7

i need the answer asap plz A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of

her so she applies the brakes and slows to 8.0 km/hr in 0.25 sec. What is the acceleration of the bike rider during this period? 1.0 km/hr/sec -1.0 km/hr/sec 16 km/hr/sec -16 km/hr/sec
Physics
2 answers:
Kruka [31]3 years ago
5 0
-16 km/hr/sec because she slows down 4 km/hr (12 - 8) in 0.25 seconds. in one second (0.25 sec * 4) she would slow down by 16 km/hr (4 * 4). it’s negative acceleration because she’s slowing down
Bas_tet [7]3 years ago
3 0

Answer:      Acceleration of the bike rider during this period is   -16 km/hr/s

Explanation:Average acceleration is equal to change in velocity divided by time taken

therefore, a=\frac{Vf-Vi}{t}

=> a=\frac{8-12}{0.25} km/hr/s = -16 km/hr/s

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A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly
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Answer:

a. B= 9.45 \times10^{-3} T

b. B= 0.820 T

c. B= 0.0584 T

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

J= \frac {-I_{c}}{\pi(c^{2}-b^{2}  ) }}

J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}

B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\

B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

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